Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N;
pair<int, int> arr[200005];
int pre[200005];
int bs(long double fnd) {
  int l = 1, r = N, ret = 0;
  while (l <= r) {
    int mid = l + r >> 1;
    if (arr[mid].second >= fnd) {
      ret = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(10);
  int by, xl, xr;
  cin >> by >> xl >> xr;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    int l, r;
    cin >> l >> r;
    arr[i] = make_pair(l, r);
    pre[i] = pre[i - 1] + r - l;
  }
  arr[N + 1].first = 1100000000;
  int Q;
  cin >> Q;
  while (Q--) {
    int x, y;
    cin >> x >> y;
    long double slope = 1.0 * (by - y) / (xl - x);
    long double tar = (slope * x - y) / slope;
    if (xl == x) {
      tar = x;
    }
    int lft = bs(tar);
    long double out = -pre[lft];
    if (tar <= arr[lft].first) {
      out += arr[lft].second - arr[lft].first;
    } else {
      out += arr[lft].second - tar;
    }
    slope = 1.0 * (by - y) / (xr - x);
    tar = (slope * x - y) / slope;
    if (xr == x) {
      tar = x;
    }
    int rht = bs(tar);
    if (arr[N].second < tar) {
      rht = N + 1;
    }
    if (rht == 0) {
      cout << 0 << "\n";
    } else {
      out += pre[rht - 1];
      out += max(0.0, (double)(tar - arr[rht].first));
      cout << max(0.0, (double)((out * (y - by)) / y)) << "\n";
    }
  }
}